Aptitude - Calendar - Discussion
Discussion Forum : Calendar - General Questions (Q.No. 2)
2.
What was the day of the week on 28th May, 2006?
Answer: Option
Explanation:
28 May, 2006 = (2005 years + Period from 1.1.2006 to 28.5.2006)
Odd days in 1600 years = 0
Odd days in 400 years = 0
5 years = (4 ordinary years + 1 leap year) = (4 x 1 + 1 x 2) 
6 odd days
Jan. Feb. March April May (31 + 28 + 31 + 30 + 28 ) = 148 days
148 days = (21 weeks + 1 day) 1 odd day.
Total number of odd days = (0 + 0 + 6 + 1) = 7 0 odd day.
Given day is Sunday.
Discussion:
214 comments Page 18 of 22.
Kishor said:
8 years ago
Well explained @Meena.
Hrithik said:
8 years ago
Thanks for the explanation @Rajat.
Neenu said:
8 years ago
Hi, I am not understanding the steps. Can anybody help me? please.
Nirmit said:
8 years ago
In every 400 years, there are 303 non-leap years and 97 leap years. ( 400/4 - 3 = 97 )
How? Every fourth year is a leap year, but century years (i.e, ending with 00) are considered leap years only if they are divisible by 400. So every 100th,200th and 300th years are non- leap years and the 400th year is a leap year.
The Number of odd days in every year= 365%7=1.
The Number of odd days in every leap year= 366%7=2.
The Number of odd days in every 400 years= 303*1 + 97*2 = 497.
But since 497 is a multiple of 7 (i.e, 497%7=0) , odd days is same as 0.
Now, the year 2000 has five such 400 years. So number of odd days in 2000 years= 5 * number of odd days in 400 = 5*0 = 0.
Then follow explanation for the next six years as given in solution.
Hope this helps!
How? Every fourth year is a leap year, but century years (i.e, ending with 00) are considered leap years only if they are divisible by 400. So every 100th,200th and 300th years are non- leap years and the 400th year is a leap year.
The Number of odd days in every year= 365%7=1.
The Number of odd days in every leap year= 366%7=2.
The Number of odd days in every 400 years= 303*1 + 97*2 = 497.
But since 497 is a multiple of 7 (i.e, 497%7=0) , odd days is same as 0.
Now, the year 2000 has five such 400 years. So number of odd days in 2000 years= 5 * number of odd days in 400 = 5*0 = 0.
Then follow explanation for the next six years as given in solution.
Hope this helps!
S.kowsalya said:
8 years ago
What was the day in 22 march, 1989? Can anyone solve this?
Suresh said:
8 years ago
Can I do this way?
28+2+1+1=32.
32÷7=remains is 4.
So the answer is,
Because in 2000-2100.
Week of days.,
2000,jan1 is sat.
So,
Fri. Sat. Sun. Mon Tue Wed Thrus
1. 2. 3. 4. 5. 6. 0
Ans is Monday.
28+2+1+1=32.
32÷7=remains is 4.
So the answer is,
Because in 2000-2100.
Week of days.,
2000,jan1 is sat.
So,
Fri. Sat. Sun. Mon Tue Wed Thrus
1. 2. 3. 4. 5. 6. 0
Ans is Monday.
Priya said:
8 years ago
Simple as it is,
Calculate the odd days from 2000 to 2006.
2004 is a leap year has 2 odd days.
(00+01+02+03+04+05+06) years = 1+1+1+1+2+1 odd days respectively since 2006 has it february.
= 7 % (mod) 7 = 0.
Which is sunday.
Calculate the odd days from 2000 to 2006.
2004 is a leap year has 2 odd days.
(00+01+02+03+04+05+06) years = 1+1+1+1+2+1 odd days respectively since 2006 has it february.
= 7 % (mod) 7 = 0.
Which is sunday.
Heeth said:
8 years ago
Thank you so much @Deena.
Rahu said:
8 years ago
Can we take 2000 instead of 1600 and 400?
Md Ajmal Pasha said:
8 years ago
Thanks so much @Deena.
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